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| sum https://forumdematematica.org/viewtopic.php?f=22&t=299 |
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| Autor: | kinu [ 13 abr 2012, 10:07 ] |
| Título da Pergunta: | sum |
find the sum of \(n\) terms of the series \(1+2(1-a) +3(1-a)(1-2a)..........+k(1-a)(1-2a)...\{1-(k-1)a\}\) |
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| Autor: | Rui Carpentier [ 17 abr 2012, 18:48 ] |
| Título da Pergunta: | Re: sum |
Well, another hard work problem. I'm not gonna do it, just give you some hints: Let \(S(a)=1+2(1-a)+3(1-2a)+\cdots + n(1-a)(1-2a)\{1-(n-1)a\}\), that is \(S(a)=\sum_{k=1}^{n-1}k\prod_{i=1}^{k-1}(1-ia)\). 1. Show that \(S\) is a polynomial of degree \(n-1\) over \(a\). (Easy) 2. Show that, for \(k=1, 2,\dots ,n\), \(S(\frac{1}{k})=k\). (It is hard, my suggestion is start the sum by the end.) 3. Find \(S(a)\) by using polynomial interpolation. You can use the inverse of the Vandermonde matrix or the Lagrange polynomials. (I didn't do it and I'm not sure that the final answer will be more simple than the inicial sum. In fact, it seems to me that original sum is the polynomial obtained by Newton interpolation of \((1,1), (\frac{1}{2},2),\dots, (\frac{1}{n},n)\).) |
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| Autor: | kinu [ 23 abr 2012, 03:35 ] |
| Título da Pergunta: | Re: sum |
Thanks Rui for precious hint |
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